Question

In figure the side ab and ac of triangle ABC are produced to point a and D respectively if bisector of B and C of angle c b and b c d respectively meet at point O then prove that angle BOC equal to 90 degree minus half of angle BAC ok

Answer 1

∠CBE = 180 - ∠ABC

∠CBO=1/2∠CBE(BO is the bisector of ∠CBE)

∠CBO = 1/2 ( 180 - ∠ABC) 1/2 x 180 = 90

∠CBO = 90 - 1/2 ∠ABC.....(1)

1/2×∠ABC = 1/2∠ABC

∠BCD = 180 - ∠ACD

∠BCO = 1/2 ∠BCD(CO is the bisector of ∠BCD)

∠BCO = 1/2 (180 - ∠ACD)

∠BCO = 90 - 1/2∠ACD......(2)

∠BOC = 180 - (∠CBO + ∠BCO)

∠BOC = 180 - (90 - 1/2∠ABC + 90 - 1/2∠ACD)

∠BOC = 180 - 180 + 1/2∠ABC + 1/2∠ACD

∠BOC = 1/2 (∠ABC + ∠ACD)

∠BOC = 1/2 ( 180 - ∠BAC) (180 -∠BAC = ∠ABC + ∠ACD)

∠BOC = 90 - 1/2∠BAC

Hence proved

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